What is the integral from 0 to 2 of sqrt(16-2x) ?

The integral from 0 to 2 of sqrt(16-2x) is (8(8-3sqrt(3)))/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2 of sqrt(16-2x)

\int_{0}^{2} 16 - 2 x d x

\frac{8 ( 8 - 3 3 )}{3}

Decimal

7.47692 \dots

Solution steps

\int_{0}^{2} 16 - 2 x d x

Apply u-substitution

: \int_{16}^{12} - \frac{u}{2} d u

= \int_{16}^{12} - \frac{u}{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{12}^{16} - \frac{u}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \frac{1}{2} \cdot \int_{12}^{16} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{12}^{16}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{12}^{16})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{12}^{16}

Compute the boundaries:

\frac{128}{3} - 163

= \frac{1}{2} (\frac{128}{3} - 163)

Simplify

= \frac{8 ( 8 - 3 3 )}{3}

\int_{- 1}^{1} ln (x + 2) d x

\int_{0}^{\frac{π}{2}} \frac{1}{cos ^{2} ( x )} d x

\int_{0}^{\frac{π}{4}} cos (x) d x

\int_{- 2}^{2} x (x^{2} - 7) d x

\int_{0}^{π} e^{x} cos^{2} (x) d x

What is the integral from 0 to 2 of sqrt(16-2x) ?
The integral from 0 to 2 of sqrt(16-2x) is (8(8-3sqrt(3)))/3